By Prof. Lefteris Kaliambos (Λευτέρης Καλιαμπός) Τ.Ε. Institute of Larissa Greece.

( June 2014)

After the discovery of the assumed uncharged neutron (1932) and the invalid relativity (1905) which led to the abandonment of the well-established electromagnetic laws, theoretical physicists developed fallacious nuclear theories for the nuclear force and various nuclear structure models, which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of nine extra charged quarks in proton and nine ones in neutron able to give the nuclear binding and nuclear structure. (See my papers in my FUNDAMENTAL PHYSICS CONCEPTS ).

Nevertheless today physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications.

**Beryllium**(**Be**) has 12 known **isotopes**, but only one of these isotopes (9Be) is stable and a primordial nuclide. As such, beryllium is considered a monoisotopic element. It is also a mononuclidic element, because its other isotopes have such short half-lives that none are primordial and their abundance is very low. Beryllium is unique as being the only monoisotopic element with both an even number of protons and an odd number of neutrons. Here we describe the ubstable Be8 and the stable Be9.

To simplify the problem you can use the following diagrams of Be-8 and Be-9, and for detailed knowledge you can see Fig. 5a and Fig. 5b of my published paper.

**NUCLEAR STRUCTURE OF Be-8 WITH S =0**

To compare the structure of Be-8 with the structures of He-4, O-16, and Pb-208 see the following figures.

Also in the following first diagram of Be-8 you see that it consists of the two simple rectangles (p_{1}n_{1}p_{2}n_{2}) and (p_{3}n_{3}p_{4}n_{4}) which represent the rectangles of two alpha particles having a total binding energy 2B(He4) = 2(-28.29) MeV with S = 0. However when the two alpha particles are closely packed for the formation of the unstable parallelepiped of Be-8 one observes that two extra pp repulsions as p_{1}p_{3} = p_{2}p_{4} appear. They are very strong because the protons have parallel spin and exert both electric and magnetic repulsions along the diagonals of the squares (p_{1}n_{1}p_{3}n_{3}) and (p_{2}n_{2}p_{4}n_{4}) respectively. Note that the nucleons cannot form a cube because of their oblate spheroid. In the same way one sees the nn repulsions as n_{1}n_{3} = n_{2}n_{4}. Also between the two alpha particles along the diagonals of the rectangles (n_{1}p_{3}n_{4}p_{2}) and (p_{1}n_{3}p_{4}n_{2}) one observes the repulsions p_{2}p_{3} = p_{1}p_{4} and n_{1}n_{4} = n_{2}n_{3} respectively. All these repulsions give a total extra repulsive energy U_{ex} as

U_{ex} = 2U(p_{1}p_{3}) + 2U(p_{2}p_{3}) + 2U (n_{1}n_{3}) +2U(n_{2}n_{3})

Of course such extra repulsions contribute to the reduction of the binding energy of the four extra bonds having the structure of deuterons, as p_{1}n_{3} = n_{1}p_{3 }= p_{2}n_{4} = n_{2}p_{4} with a total extra binding energy

B_{ex} = 4B(p_{1}n_{3}) .

**Diagram of** **stable Be-9 with S =-3/2**

** n _{4 }(-1/2)..p_{4}( -1/2)..n_{5}(-1/2) **

** p _{2}(+1/2)..n_{3 }(+1/2).p_{3}(+1/2) **

** '*** '*** n1( -1/2)..p _{1}(-1/2)..n_{2}( -1/2) **

** ** ** **

'* '*

**'**

*'*

**DIAGRAM OF THE UNSTABLE Be-8 WITH S =0**

Here the p_{1}n_{1}n_{2}p_{2} make the first horizontal square with positive spins at the first horizontal plane (+HP1) , while the n_{3}p_{3}p_{4}n_{4} make the second horizontal square at the second horizontal plane of negative spins (-HP2).All these nucleons form the first unstable parallelepiped of the structure of atomic nuclei.

** p _{4}.........n_{4}**

** n _{3}.........p_{3} -HP2**

** n _{2}..........p_{2}**

**p _{1}..........n_{1} +HP1 **

Using the binding energies B(Be-8) = -56.5 MeV and B( He-4) = -28.29 MeV one can write

B(Be-8) = 2B(He-4) + ( U_{ex} + B_{ex}) Or -56.5 = 2(-28.29) + ( U_{ex} + B_{ex})

That is ( U_{ex} + B_{ex}) = 0.08 MeV

It means that the large number of extra repulsions overcome the extra bonds and lead to the decay of Be8 which splits into two alpha particles. It is of interest to note that the fallacious nuclear structure model of the well-known Fermi gas leads to serious problems, because one might expect that Be8 with a great symmetry with S = 0 and Z = N could be able to be one of the most stable nuclides. Actually the Be8 of a great symmetry is the collection of two very stable alpha particles (rectangles) which cannot form a stable nucleus in three dimensions for making the first parallelepiped in the nuclear structure.

**NUCLEAR STRUCTURE OF Be-9 WITH S =-3/2**

In the stable Be-9 with S = -3/2 as shown in the second diagram the structure occurs not as a parallelepiped in three dimensions but as a rectangle (n_{1}n_{2}n_{5}n_{4}) in two dimensions in which a smaller rectangle (n_{1}p_{1}p_{4}n_{4}) could represent the nuclear structure of Li-6. In this case one observes that outside of Li-6 there are extra repulsions and extra pn bonds . The extra pp repulsions as p_{3}p_{1}, p_{3}p_{2} and p_{3}p_{4} have an extra U(pp) repulsive energy, while the extra nn repulsions as n_{2}n_{1}, n_{2}n_{3}, n_{2}n_{4}, and n_{2}n_{5}, along with the extra repulsions as n_{5}n_{1}, n_{5}n_{3} and n_{5}n_{4} have a total extra U(nn) repulsive energy. Here one concludes that all these extra repulsive energies U_{ex} = U(pp) + U(nn) cannot overcome the extra total binding energy B_{ex} _{ }of the pn bonds like p_{1}n_{2}, p_{3}n_{3}, p_{4}n_{5,} p_{3}n_{2}, and p_{3}n_{5 , }because the p_{3}n_{2} and p_{3}n_{5 }systems are very strong bonds acting along the spin axis.

Using the binding energies B(Be-9) = -58.15 MeV and B(Li-6) = -31.98 MeV one can write

B(Be-9) = B(Li-6) + (B_{ex} + U_{ex} ) Or -58.15 = -31.98 + (B_{ex} +U_{ex})

That is (Bex + Uex) = - 26.17 MeV

This means that the radial and the strong axial pn bonds overcome the pp and nn repulsions and lead to the stability of Be-9. Note that nuclear physicist applying not the electromagnetic laws but the fallacious nuclear structure models believe that Be-9 is composed of two α-particles separated to form a deformed Be8-core and an extra neutron strongly coupled to the motion of the Be8-core by the neutron core potential. ( See “Alpha-Particle Model for Be9 – Progress of Theoretical Physics ”).